<p>Let's make this the official thread to discusss answers to the FRQ.</p>
<p>hey i have my frq stuff back. if anyone wnats to IM me and compare answers feel free.</p>
<p>xonicole143ox.</p>
<p>Wooooooooooooooooooooooooooooooooooooooo FR discusssiion</p>
<p>although i dont really remember the questions..........</p>
<p>I don't remember my answers specifically (which I probably will remember in at least 50 minutes when TCB posts the FR questions and I do them again), but for the first problem, I recall getting concentrations that were exactly the same as the concentrations given in the problem.</p>
<p>For problem 3, I think I got C2N5H4O, something like that. As for the second part of that problem, I multiplied the subscripts by two.</p>
<p>I don't remember all of the equations that I did.</p>
<p>You know what, at this point, I'm not going to say anything else because I pretty much forget all specific answers. I'll just wait until I redo the FR questions within a few hours.</p>
<p>i got the same for #3, 6570882.</p>
<p>What are your answers for the equilibrium problem? The questions are online now. </p>
<p>did anyone else fail to count how many rows down the periodic table Q would be in?</p>
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<p>_></p>
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<p>a)ii: Pb2+ = 1.3<em>10^-3, I1- = 2.6</em>10^-3 M
iii: 8.8<em>10^-9
b) Double the [Pb2+] from above
c) [Pb2+] decreases
d) i: 4.2</em>10^-6 mol for both
ii:the Ksp expression using these concentrations gives 1.8<em>10^-11 mol, less than the 1.2</em>10^-10 needed to form a precipitate.</p>
<p>ha, i put 1s1 for the configuration instead of 8s1. hahah. i know people that sat there and wrote the whole thing.</p>
<p>kman I got that!</p>
<p>That first problem seemed dubiously too easy.</p>
<p>Your B is wrong. Increasing volume of the PbI2 has no effect on the other two things. (Pb and I)</p>
<p>It was in two liters, not one, and it was to make a saturated solution, not adding PbI2 to a saturated solution.</p>
<p>What are the answers for 1,2,4,5,6,and 8?</p>
<p>yea i said in b, it had no effect</p>
<p>Me too, otherwise I got your answers, kman.</p>
<p>I did these reactions:</p>
<p>1) KClO4 --> KCl + KClO3</p>
<p>2) C6H14 + O2 --> CO2 + H2O</p>
<p>3) NH3 + HF --> NH4+ + F-</p>
<p>4) CH3COOH + OH- --> HOH + CH3COO-</p>
<p>5) Zn + Cu2+ --> Zn2+ + Cu</p>
<p>I can give the answers for 7, since that was the easiest of all the problems.
..
:F:
| ..
: S - F::
|
:F::</p>
<p>sp³ hybridization
trigonal pyramidal
Less than 109.5º because that lone pair of electrons displaces electron density around the other domains, pushing more forcefully on those angles.</p>
<p>.. ..
:F: :F:
| / ..
: S -- F::
| \ ..
:F:: :F:</p>
<p>sp³d² hybridization
square pyramidal
S is +4.</p>
<p>oh poop, it messed up my spacing in those Lewis structures. Just imagine octets around each F and an extra : on each S.</p>
<p>I see-one dissolves 2.6*10^-3 mol Pb2+ in 2 L of solution, so the M remains unchanged even though the #moles is doubled. I stand corrected.</p>